User:Ageary/Chapter 6

Problems in Exploration Seismology and their Solutions

Problems-in-Exploration-Seismology-and-their-Solutions.jpg
Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

6.1 Characteristics of different types of events and noise

Classify the different types of events and noise in Table 6.1a on the basis of commonly observed characteristics on a seismic record.

Figure 6.1a.  Characteristics of events.

Figure 6.1b.  Characteristics of events.

Background

A wave is coherent when it appears on successive traces in a systematic way and with approximately the same waveshape. Apparent-velocity filtering, also called apparent-dip filtering, refers to discrimination (attenuation) based on the slope of a linear alignment of traces across a section (see problem 9.25). Frequency filtering refers to attenuation of certain frequencies relative to other frequencies (see problems 7.11 and Sheriff and Geldart, 1995, Section 9.5.10). Arrays, discussed in Sheriff and Geldart (1995, Section 8.3.5 to Section 8.3.9) discriminate on the basis of apparent wavelength. Common-midpoint (CMP) stacking is discussed in Sheriff and Geldart, 1995, Section 8.3.3. Muting involves setting to zero the parts of traces prior to a certain ‘mute schedule.’ Three-component recording uses geophones that respond to motion along three different orthogonal axes, e.g., north-south, east-west, and vertical.

Solution

In Table 6.1b we assume that the dimensions are those commonly used, that “distinctive apparent velocity” means apparent velocity within certain limits, that structural and other changes are generally small, that the 3-component phones are on bedrock rather than on top of a low-velocity layer. Usually only wind noise and other nearly random background noises are incoherent and all source-generated events are predictable trace-to-trace and repeatable. CMP stacking should attenuate events that do not have the same hyperbolic relationship (see problem 4.1a) and stacking velocity as the primary reflections. In Table 6.1b, (1) indicates that the attenuation will be mainly that of the diffraction tails, (2) indicates that reflected refractions from off to the side of the line may have hyperbolic curvature, and (3) indicates that Love waves are dispersive.

6.2 Horizontal resolution

Assume that a salt dome can be approximated by a vertical circular cylinder with a flat top of radius 400 m at a depth of 3200 m. If the average velocity above the top is 3800 m/s, what is the minimum frequency that will give a recognizable reflection from the dome?

Background

Huygens’s principle (see problem 3.1) states that waves are reflected from all illuminated parts of a reflector, the phase varying with the two-way traveltime from source to reflecting point to receiver. Thus the receiver records energy from all points of the reflecting area, the “reflection” being the sum of all of the increments, each with a different phase.

The first Fresnel zone (often referred to as “the Fresnel zone”) is the portion of the reflector from which the reflected energy arrives more-or-less in-phase so that it adds constructively. For constant velocity, it is a circle centered at the reflecting point $P_{o}$ and extending out to where the slant distance $h_{1}$ is such that $h_{1}=h_{0}+\lambda /4$ (see Figure 6.2a). Because $h_{0}\gg \lambda /4$, the Fresnel zone radius $R$ is

${\begin{aligned}R=(h_{1}^{2}-h_{0}^{2})^{1/2}\approx (\lambda h_{0}/2)^{1/2}.\end{aligned}}$

(6.2a)

The annular ring defined by $h_{1}$ and $h_{2}$, where $h_{2}=h_{1}+\lambda /4=h_{0}+2(\lambda /4)$, is the second Fresnel zone, the outer radius being $R_{2}=(\lambda h_{0})^{1/2}$, and so on for successive zones. The amplitude of the total reflected energy as a function of $R$ is plotted in Figure 6.2b (see Sheriff and Geldart, 1995, section 6.2.3 for more details). The amplitude depends mainly on the first zone, the contributions of successive pairs of the other zones effectively cancelling each other. The first Fresnel zone is usually taken as the limit of the horizontal resolution for unmigrated seismic data, reflectors smaller than this appearing almost as point diffractors.

Aliasing is discussed in problem 9.4.

Figure 6.2a.  Geometry of first Fresnel zone.

Figure 6.2b.  Amplitude versus reflector radius.

Solution

For a recognizable reflection (as opposed to a diffraction) on an unmigrated section, the radius of the dome should be at least as large as that of the first Fresnel zone, that is,

${\begin{aligned}R>(\lambda h_{0}/2)^{1/2}=(Vh_{0}/2f)^{1/2}.\end{aligned}}$

Solving for the frequency $f$, we have

${\begin{aligned}f=Vh_{0}/2R^{2}=3.8\times 3.2/(2\times 0.4^{2})=38\ {\rm {Hz}}.\end{aligned}}$

For frequencies lower than 38 Hz, the top of the dome is smaller than the Fresnel zone and the reflection energy falls off so that the reflection may not be recognized as such.

6.3 Reflection and refraction laws and Fermat’s principle

6.3a Use Fermat’s principle of stationary time to derive the law of reflection

Background

In the solution of problem 3.1a we showed that the angle of incidence equals the angle of reflection and that, for the angle of refraction $\theta _{2}$, ${\rm {\;sin\;}}\theta _{2}=(V_{2}/V_{1}){\rm {\;sin\;}}\theta _{1}$ [see also equation (3.1a)].

These are the laws of reflection and refraction. Fermat’s principle of least time (more accurately, of stationary time) states that wave travel between any two points is along the path for which the traveltime is either a maximum or a minimum value (i.e., the derivative of the traveltime equals zero) compared with the traveltimes along adjacent paths.

Figure 6.3a.  Deriving Snell’s law.

Solution

In Figure 6.3a, the source $S$ and the receiver $R$ have coordinates $(0,\;h_{1})$ and $(a,\;h_{2})$. The traveltime for a wave from $S$ to $R$ with reflecting point $M(x,\;0)$ is

${\begin{aligned}t=(1/V)\{(x^{2}+h_{1}^{2})^{1/2}+[(a-x)^{2}+h_{2}^{2}]^{1/2}\}.\end{aligned}}$

To find the point $M$ for which the value of $t$ is stationary, we differentiate $t$ with respect to $x$ and set the result equal to zero. Thus,

${\begin{aligned}{\frac {{\rm {d}}t}{{\rm {d}}x}}=(1/V)\left\{{\frac {x}{(x^{2}+h_{1}^{2})^{1/2}}}-{\frac {(a-x)}{[(a-x)^{2}+h_{2}^{2}]^{1/2}}}\right\}=0.\end{aligned}}$

The two terms in the brackets are the sines of the angles $\theta _{1}$ and $\theta _{1}^{'}$; hence, ${\rm {\;sin\;}}\theta _{1}={\rm {\;sin\;}}\theta _{2}^{'}.$

6.3b Repeat part (a) for the refracted path SMQ, in Figure 5.3a.

Solution

The traveltime for the path SMQ is

${\begin{aligned}t=(x^{2}+h_{1}^{2})^{1/2}/V_{1}+[(b-x)^{2}+h_{3}^{2}]^{1/2}/V_{2}.\end{aligned}}$

Differentiation gives

${\begin{aligned}{\frac {{\rm {d}}t}{{\rm {d}}x}}={\frac {x}{V_{1}(x^{2}+h_{1}^{2})^{1/2}}}-{\frac {(b-x)}{V_{2}[(b-x)^{2}+h_{3}^{2}]^{1/2}}}=0,\end{aligned}}$

that is, ${\rm {\;sin\;}}\theta _{1}/V_{1}={\rm {\;sin\;}}\theta _{2}/V_{2}$.

6.3c Repeat parts (a), (b) for reflected and refracted converted S-waves.

Solution

If we replace the angles $\theta$ with the angles $\delta$ and use the S-wave velocities $\beta$, the foregoing proofs are otherwise unchanged.

6.4 Effect of reflector curvature on a plane wave

Redraw Figure 6.4a for a plane wave incident on the reflector, and explain the significance of the changes which this makes.

Background

Figure 6.4a assumes a point source at the surface, whereas for an incident plane wave in Figure 6.4b the source is at infinity. The plane wave reflected by the plane reflector produces a plane wavefront (R), i.e., the reflected wavefront has zero curvature. For a point diffractor the virtual source is at the diffractor and the wavefront has the maximum curvature (D). The curvature of the wavefront from the anticlinal reflector (A) is intermediate between those of $R$ and $D$ and the curvature of the wavefront from the synclinal reflector (S) is negative (assuming the center of curvature is below the surface).

Figure 6.4a.  Reflector curvature effects for point source.

Figure 6.4b.  Reflection curvature for incident plane wave.

Solution

By Huygens’s principle (problem 3.1), a diffracting point acts as a point source whenever a wave falls upon it; hence, the diffraction response $D$ to a plane wave (Figure 6.4b) is the same as that in Figure 6.4a. A plane wave incident on a plane reflector gives rise to a reflected plane wave $R$. For a plane wave incident on an anticline or syncline of circular cross-section of radius $r$, we can use the mirror formula, namely,

${\begin{aligned}1/f=2/r=1/u+1/v,\end{aligned}}$

where $f$ is the focal length $=r/2$, $u$ and $v$ are the distances of the source and reflected image from the apex of an anticline or trough of a syncline; $r$ is positive for a syncline and $u$ is infinite for a plane wave, so $v=r/2$. For a syncline, the reflected wave comes to a focus a distance $r/2$ above the trough. For an anticline, a virtual image (see problem 4.1) is at $r/2$ below the high point.

Figure 6.5a.  Diffraction traveltime curves.

6.5 Diffraction traveltime curves

6.5a Show that the slope of the diffraction curve with source $S_{2}$ in Figure 6.5a(i) approaches $\pm 1/V$ for large $x$.

Solution

The diffraction path is $S_{2}AG_{2}$ in Figure 6.5a(i), so the traveltime curve is

${\begin{aligned}t_{d}=h/V+(x^{2}+h^{2})^{1/2}/V=h/V+(x/V)[1+(h/x)^{2}]^{1/2}.\end{aligned}}$

For $x\gg h$, the equation of the curve becomes

${\begin{aligned}t_{d}\approx x/V+h/V\end{aligned}}$

which is a straight line with slope $+1/V$ for $x>0$,$-1/V$ for $x<0$. The traveltime approaches these asymptotes as $x\to \pm \infty .$

Alternative solution

The slope of the traveltime curve is

${\begin{aligned}{\frac {{\rm {d}}t_{d}}{{\rm {d}}x}}={\frac {x}{V(x^{2}+h^{2})^{1/2}}}=\pm {\frac {1}{V}}[1+(h/x)^{2}]^{-1/2}.\end{aligned}}$

For $|x|\gg h$, the slope is $\pm 1/V$ as before.

6.5b What is the asymptote slope for a coincident source-receiver?

Solution

The traveltime curve for Figure 6.5a(ii) is given by

${\begin{aligned}t_{d}=(2/V)(x^{2}+h^{2})^{1/2}=(\pm 2x/V)[1+(h/x)^{2}]^{1/2}\end{aligned}}$

${\begin{aligned}\approx (\pm \ 2x/V)[1+{\frac {1}{2}}(h/x_{1})^{2}].\end{aligned}}$

As $|x|$ increases, $t_{d}\to \pm 2x/V$. The asymptote has the equation $t_{d}=\pm 2x/V$, which is a straight line with slope $\pm 2/V.$

6.6 Amplitude variation with offset for seafloor multiples

6.6a Assume that Figure 6.6a shows relative amplitudes correctly (divergence having been allowed for). The water depth is 0.42 km and the velocity below the seafloor is 2.59 km/s. If the reflection coefficient is maximum at the critical angle, on which traces would you expect the maximum amplitude for the first, second, third, and fourth multiples?

Background

Divergence or geometrical spreading and multiples were discussed in problem 3.8, the reflection coefficient (reflectivity) $R$ in problem 3.6, the critical angle in problem 4.18. A refraction does not exist to the left of $Q$ (in Figure 4.18a); $OQ=2h{\rm {\;tan\;}}\theta _{c}=x^{'},x^{'}$ being the critical distance. The reflectivity $R$ (see equation 3.6a) is generally maximum when $\theta =\theta _{c}$ (see Sheriff and Geldart, 1995, section 3.3).

Figure 6.6a.  Record showing seafloor multiples.

Solution

We interpret the five more-or-less equally spaced events in Figure 6.6a as a primary reflection followed by four multiples. We assume that the amplitude maxima occur when the angle of incidence equals the critical angle. Hence the offsets for the amplitude maxima correspond to multiples of the critical distance.

Taking the water velocity as 1.50 km/s, $\theta _{c}={\rm {sin}}^{-1}(1.50/2.59)=35.4^{\circ }$. If we assume that the source and receiver are both at depths of 15 m (the conventional streamer depth), the critical distance $x^{'}$ is

${\begin{aligned}x^{'}=2z{\rm {\;tan\;}}\theta _{c}=2(420-15){\rm {\;tan\;}}35.4^{0}=580\ {\rm {m}}.\end{aligned}}$

The maximum amplitudes occur at offsets of approximately 600, 1200, 1800, 2400, and 3000 m, that is, at 600 m intervals, close to the calculated critical distance.

6.6b What should be the ratio of the amplitudes of successive multiples on the short-offset trace? How do these calculations compare with observations? What unaccounted for factors affect this comparison?

Solution

The ratio of successive amplitudes should be $-R$. Taking the amplitude of the incident wave as unity, the amplitudes of the multiples become $-R$, $R^{2}$, $-R^{3}$, $R^{4}$. The measured amplitudes on the shortest traces are 32 (at offset 425 m), 12, 4.5, 2.0, 1.2 mm, the first two measurements not being very accurate. The ratios of the successive measurements are 2.7, 2.7, 2.2, 1.7, the average value being about 2.3. Assuming the value of 2.3,

${\begin{aligned}R=1/2.3=0.43=(Z_{2}-Z_{1})/(Z_{2}+Z_{1}),\end{aligned}}$

where $Z_{t}=\rho _{i}V_{i}$ (see problem 3.6). Taking $\rho _{1}=1.03\ {\hbox{g}}/{\hbox{cm}}^{3}$, $Z_{1}=1.03\times 1.50=1.54$, we get $Z_{2}=3.9$. Since $V_{2}=2.59\ {\rm {km/s}}$, we have only one unknown, $\rho _{2}$, and we get $\rho _{2}\approx 1.5$ ${\rm {g/cm}}^{3}$, a reasonable value (but nevertheless of questionable accuracy in view of the many uncertainties involved).

The above comparison would be affected principally by time-dependent factors that have not been taken into account, such as absorption, transmission losses, peg-leg multiples (problem 3.8), etc. Losses of these kinds are very small for wave travel in water. Amplitude changes may also have occurred during data processing.

6.7 Ghost amplitude and energy

6.7a If the source depth is $c\lambda$ (where $\lambda$ is the wavelength) and $0<c<+1$ in equation (6.7a), discuss the conditions under which the amplitude of $\psi _{P}$ is zero.

Background

The low-velocity layer (LVL) is discussed in problem 4.16, ghosts in problem 3.8. When a ghost is superimposed on a downgoing wave, it affects not only the waveshape but also the directivity. In Figure 6.7a, $S$ is a point source at a depth $c\lambda$ and $I$ is the image point (see problem 4.1) for energy reflected at the surface. For a ghost arriving at point $P$, the virtual path is $IP$. If the source emits the wave $\psi _{P}=A{\rm {\;cos\;}}(\kappa r-\omega t)$ and the reflection coefficient at the surface is $-1$, the combined primary wave plus ghost at point $P$ is

${\begin{aligned}\psi _{P}=A{\rm {\;cos\;}}(\kappa r_{1}-\omega t)-A{\rm {\;cos\;}}(\kappa r_{2}-\omega t).\end{aligned}}$

Using the identity ${\rm {\;cos\;}}x-{\rm {\;cos\;}}y=-2{\rm {\;sin\;}}[(x+y)/2]{\rm {\;sin\;}}[(x-y)/2]$, we get

${\begin{aligned}\psi _{P}=-2A{\rm {\;sin\;}}[\kappa (r_{1}+r_{2})/2-\omega t]{\rm {\;sin\;}}[\kappa (r_{1}-r_{2})/2].\end{aligned}}$

When $r\gg c\lambda$, $r_{1}\approx r-c\lambda {\rm {\;cos\;}}\theta$, $r_{2}\approx r+c\lambda {\rm {\;cos\;}}\theta$; also $c\lambda =2\pi c/\kappa$, so we get

${\begin{aligned}\psi _{P}=2A{\rm {\;sin\;}}(\kappa r-\omega t){\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta )\\\qquad =[2A{\rm {\;sin\;}}(2\pi c\ {\rm {cos}}\ \theta )]{\rm {\;cos\;}}(\kappa r-\omega t-\pi /2).\end{aligned}}$

(6.7a)

Transmissivities $T\uparrow$ and $T\downarrow$ are defined in problem 3.6 where equation (3.6c) shows that

${\begin{aligned}T\uparrow +T\downarrow =2,\;T\uparrow T\downarrow =E_{T}.\end{aligned}}$

(6.7b)

Absorption is discussed in problem 2.18.

Figure 6.7a.  Directivity of source plus ghost.

Solution

Equation (6.7a) gives for the amplitude of the primary wave plus ghost,

${\begin{aligned}A^{*}=2A{\rm {\;sin\;}}(2\pi c\ {\rm {cos}}\ \theta ).\end{aligned}}$

(6.7c)

For $A^{*}=0$, $2\pi c{\rm {\;cos\;}}\theta =n\pi$, that is, ${\rm {\;cos\;}}\theta =n/2c,n=0,\pm 1,\pm 2$, .... For $n=0$, ${\rm {\;cos\;}}\theta =0$, i. e., $\theta =\pm 90^{\circ }$, and the waves are traveling horizontally. When $|n|\geq 2$ and $c<1$, ${\rm {\;cos\;}}\theta >1$ so there are no appropriate values of $\theta$.

6.7b For a source below the base of the LVL, compare the amplitude and energy of ghosts generated at the base of the LVL and at the surface of the ground, given that the velocities and densities just below and within the LVL are $V_{H}=1.9\ {\rm {km/s}}$, $\rho _{H}=2.0\ {\rm {g/cm}}^{3}$, $V_{W}=0.40\ {\rm {km/s}}$ and $\rho _{W}=1.6\ {\rm {g/cm}}^{3}$, respectively.

Solution

We assume small incidence angles so that equations (3.6a,b) are valid. Then

${\begin{aligned}Z_{1}=2.0\times 1.9=3.8;Z_{2}=0.40\times 1.6=0.64\ ({\rm {g.km/cm^{3}s}}).\end{aligned}}$

At the base of the LVL,

${\begin{aligned}R=(0.64-3.8)/(0.64+3.8)=-3.2/4.4=-0.71,\end{aligned}}$ ${\begin{aligned}E_{R}=R^{2}=0.50,\end{aligned}}$ ${\begin{aligned}T\uparrow =2\times 3.8/(3.8+0.64)=1.71\;,\;T\downarrow =0.29,\end{aligned}}$ ${\begin{aligned}E_{T}=T\uparrow \times T\downarrow =1.71\times 0.29=0.50.\end{aligned}}$

Assuming equal (unit) amplitudes for waves leaving the source in different directions, the ghost produced at the base of the LVL has amplitude $-0.71$ and energy 0.50. The ghost produced at the surface has amplitude $T\uparrow T\downarrow \times (-1)=-0.50$ and energy $E_{T}=0.50^{2}=0.25$. The amplitude of the ghost from the base of the LVL is $0.71/0.50=1.4$ times that of the ghost from the surface while the ratio of the energies is $0.50/0.25=2.4.$

6.7c Assume that the LVL is ${\frac {1}{2}}\lambda$ in thickness and that $\eta \lambda =0.6\ {\rm {dB}}$ for the LVL; now what are the ratios of the ghost amplitudes and energies?

Solution

The surface ghost has to travel a distance $\lambda$ farther than the ghost from the base of the LVL during which its amplitude is reduced by the factor $e^{-\eta x}=e^{-\eta \lambda }=e^{-0.60}=0.55$. The previous amplitude was $-0.47$, so with absorption this becomes $-0.26$, and energy becomes 0.22 $\times 0.55^{2}=0.067$. The ratios of the amplitudes and energies in part (b) now become $0.73/0.26=2.8$ and $0.53/0.067=7.9$, the dB values being 8.9 and 18.0 dB, respectively.

6.8 Directivity of a source plus its ghost

An air gun is fired at a depth of 10 m. The waveform includes frequencies in the range $10-80\ {\rm {Hz}}$, the amplitudes of the 10- and 80-Hz components being the same near the source. Compare their amplitudes for the wave plus ghost at considerable distance from the source in the directions $0^{\circ }$, $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$ to the vertical.

Background

Air guns are described in problem 7.7.

Solution

We take the velocity in water as 1.5 km/s so that the wavelengths are 150 m and 19 m for the 10-Hz and 80-Hz components. From equation (6.7a) the amplitude of the ghost is $2A{\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta )$, where the depth of the source is $c\lambda$. For the 10-Hz component, $c=10/150=0.067$; for the 80 Hz component, $c=10/19=0.53$. The ratio of the amplitude of the 10-Hz component to that of the 80-Hz component is

${\begin{aligned}{\frac {{\rm {\;sin\;}}(2\pi \times 0.067{\rm {\;cos\;}}\theta )}{{\rm {\;sin\;}}(2\pi \times 0.53{\rm {\;cos\;}}\theta )}}={\frac {{\rm {\;sin\;}}(0.42{\rm {\;cos\;}}\theta )}{{\rm {\;sin\;}}(3.3{\rm {\;cos\;}}\theta )}}.\end{aligned}}$

Table 6.8a. Ratios of ghost amplitudes of 10- and 80- Hz components.

$\theta$	0.42 ${\rm {\;cos\;}}\theta$	3.3 ${\rm {\;cos\;}}\theta$	Amplitude ratio
$0^{\circ }$	0.42	3.3	–2.6
$30^{\circ }$	0.36	2.9	1.5
$32^{\circ }$	0.30	2.3	1.04
$60^{\circ }$	0.21	1.6	0.21
$90^{\circ }$	—	—	0.13$*$

* For $\theta =90^{\circ }$, ${\rm {\;cos\;}}\theta =0$, so the ratio is 0/0. However, when $\theta$ is slightly less than $90^{\circ }$, the arguments of the sines are small and we can replace the sines with the angles; the cos factors cancel and the ratio is $0.42/3.3=0.13.$

Table 6.8a shows the results for the given values of $\theta$. The values in columns 2 and 3 are in radians and the column headed “Amplitude ratio” is the sine of the values in column two divided by the sine of the values in column three.

Thus, the 10-Hz component is stronger than the 80-Hz component as the direction approaches the vertical. The minus sign in the first ratio is due to a phase reversal of the 80-Hz component.

6.9 Directivity of a harmonic source plus ghost

Show that equation (6.7c) gives the directivity diagrams shown in Figure 6.9a.

Solution

The directivity is given by equation (6.7c). We take $2A=1$, and $c={\rm {depth}}/\lambda =0.1$, 0.5, and 1.0 for the three parts of Figure 6.9a. Then equation (6.7c) gives

${\begin{aligned}A^{*}={\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta ).\end{aligned}}$

Substituting the three values of $c$, we have:

${\begin{aligned}a):c=0.1\;,\;A^{*}={\rm {\;sin\;}}(0.63{\rm {\;cos\;}}\theta ),\end{aligned}}$ ${\begin{aligned}b):c=0.5\;,\;A^{*}={\rm {\;sin\;}}(3.1{\rm {\;cos\;}}\theta ),\end{aligned}}$ ${\begin{aligned}c):c=1.0,\;A^{*}={\rm {\;sin\;}}(6.3{\rm {\;cos\;}}\theta ).\end{aligned}}$

The results of the calculations are shown in Tables 6.9a,b.

Figure 6.9a.  Directivity of a harmonic source at depth $z=c\lambda$.

Ignoring the minus signs (which indicate phase reversals), the curves for $\varphi _{a}$ and $\varphi _{b}$, shown in Figure 6.9b, conform closely to Figure 6.9a. However, we need more points to plot the $\varphi _{c}$-curve properly and Table 6.9b shows calculated values for intermediate points. The $\Psi _{c}$-curve in Figure 6.9b also conforms closely to Figure 6.9a.

Table 6.9a. Values for $\Psi _{a}$, $\Psi _{b}$, $\Psi _{c^{'}}$.

$\theta ^{\circ }$	$\Psi _{a}$	$\Psi _{b}$	$\Psi _{c}$
0	0.59	0.00	0.00
15	0.57	0.15	−0.20
30	0.52	0.44	−0.74
45	0.43	0.81	−0.97
60	0.31	1.00	−0.01
75	0.16	0.72	1.00
90	0.00	0.00	0.00

Table 6.9b. Intermediate values for $\Psi _{c^{'}}$.

$\theta ^{\circ }$	$\Psi _{c}$	$\theta ^{\circ }$	$\Psi _{c}$
5	−0.01	50	−0.79
10	−0.08	55	−0.46
20	−0.35	65	0.46
25	−0.54	70	0.83
35	−0.90	80	0.89
40	−0.99	85	0.52

Figure 6.9b.  Calculated directivity at source depth $z=c\lambda$.

6.10 Differential moveout between primary and multiple

6.10a A multiple reflection is produced by a horizontal bed at a depth of 1100 m; average velocity to this depth is 2000 m/s. A primary reflection from a depth of 3250 m coincides with the multiple at zero offset. By how much do arrival times differ at points 200, 400, 800, and 1000 m from the source?

Solution

Raypaths for the 200 and 1000 m offsets are drawn to scale in Figure 6.10a. We treat this as a two-layer problem with a 1100-m layer over a 2150-m layer, the velocity in the second layer being such that the traveltimes in the two layers are equal. Since ${\bar {V_{1}}}=2000{\rm {m/s}}$, the travel-time $t_{0}$ in the upper layer is $2\times 1100/2000=1.100\ {\rm {s}}$ and $V_{2}=2\times 2150/1.100=3910\ {\rm {m/s}}$, the average velocity from the surface being ${\bar {V_{2}}}=2\times 3250/2.200=2950\ {\rm {m/s}}$.

Assuming a straight-line raypath at the 1000-m offset, we get an angle of ${\rm {tan}}^{-1}(500/3250)\approx 9^{\circ }$, and hence, the raypath bending will be small and we can ignore it.

The arrival time of the deep reflector is $t_{r}=(6500^{2}+x^{2})^{1/2}/2950$ and the multiple’s $t_{m}=(4400^{2}+x^{2})^{1/2}/2000$. Their arrival times and differential normal moveouts $\Delta (\Delta t_{\hbox{NMO}})$ are listed in Table 6.10a.

Figure 6.10a.  Raypaths.

6.10b If the shallow bed dips $10^{\circ }$, how much do the arrival times at 400 and 800 m change? What is the apparent dip of the multiple?

Solution

We have two cases to consider: offset updip and offset downdip. We use the notation shown in Figure 6.10b to denote the various angles of incidence and refraction. The offsets and path lengths for the shallow multiple are easily found graphically. We assume that the depths given in part (a) are vertical depths at the source. From part (a) we have $V_{2}/V_{1}=1.96$. We now calculate angles from the following relations:

${\begin{aligned}a_{1}=\xi -\alpha _{1},a_{1}^{'}={\rm {sin}}^{-1}(1.96{\rm {\;sin\;}}a_{1}),\end{aligned}}$ ${\begin{aligned}\alpha _{2}=\xi -a_{1}^{'},b_{1}^{'}=\xi +\alpha _{2}=2\xi -a_{1}^{'},\end{aligned}}$ ${\begin{aligned}\beta _{1}={\rm {sin}}^{-1}[({\rm {\;sin\;}}b_{1}^{'})/1.96].\end{aligned}}$

Table 6.10a. Arrival times and differential NMO.

$x({\hbox{m}})$	$t_{r}({\hbox{s}})$	$t_{m}({\hbox{s}})$	$\Delta (\Delta t_{\rm {NMO}})$
0	2.200	2.200	0
200	2.204	2.202	0.001
400	2.208	2.209	0.001
600	2.213	2.220	0.007
800	2.220	2.236	0.016
1000	2.229	2.256	0.027

Table 6.10b. Downdip and updip angles, offsets and traveltimes for primary.

	Downdip			Updip
$\alpha _{1}$	0.0	$1.0^{\circ }$	$2.0^{\circ }$	$4.9^{\circ }$	$7.0^{\circ }$	$8.0^{\circ }$
$a_{1}$	$10.0^{\circ }$	$9.0^{\circ }$	$8.0^{\circ }$	$5.1^{\circ }$	$3.0^{\circ }$	$2.0^{\circ }$
${a'}_{1}$	$19.9^{\circ }$	$17.9^{\circ }$	$15.8^{\circ }$	$10.0^{\circ }$	$5.9^{\circ }$	$3.9^{\circ }$
$a_{2}$	$-9.9^{\circ }$	$-7.9^{\circ }$	$-5.8^{\circ }$	$0.0^{\circ }$	$4.1^{\circ }$	$6.1^{\circ }$
$b_{2}$	$-9.9^{\circ }$	$-7.9^{\circ }$	$-5.8^{\circ }$	0.0	$4.1^{\circ }$	$6.1^{\circ }$
$b_{1}$	$0.1^{\circ }$	$+2.1^{\circ }$	$+4.2^{\circ }$	$10.0^{\circ }$	$14.1^{\circ }$	$16.1^{\circ }$
${b'}_{1}$	0.0	$+1.1^{\circ }$	$+2.1^{\circ }$	$5.1^{\circ }$	$7.1^{\circ }$	$8.1^{\circ }$
$x$	970 m	810 m	630 m	0.00	390 m	790 m
$t_{m}^{d}$	1.528 s	1.520 s	1.502 s	1.488 s	1.475 s	1.471 s

Next we plot the raypaths and measure the offsets and path lengths and finally calculate the traveltimes. The calculated angles and measured values of $x$ and $t_{r}$ are listed in Table 6.10b for the downdip and updip cases.

Figure 6.10b.  Notation for dipping bed.

The graphical construction and path length measurements are illustrated in Figure 6.10c. The primary arrival times at the required offsets are found by interpolation, using values of $x$ and $t_{r}^{d}$ in Table 6.10c; the results are shown in the first two rows of Table 6.10c. For the multiple we use the method of images (see problem 4.1) to get offsets and path lengths (see Figure 6.10c). Dividing the path lengths by the velocity gives the traveltimes in Table 6.10c.

To find how much the dip has changed the arrival times, we have inserted the zero-dip values $t_{r}$ and $t_{m}$ in Table 6.10c and entered the changes $\Delta t_{r}=t_{r}^{d}-t_{r}$, $\Delta t_{m}=t_{m}^{d}-t_{m}$.

Table 6.10c. Effect of dip on reflections.

offset	$x$	−800 m	−400	0.00	400	800
time of dipping reflection	$t_{r}^{d}$	1.520 s	1.502	1.488	1.475	1.471
time of reflection without dip	$t_{r}$	1.501 s	1.494	1.491	1.494	1.501
$t_{r}^{d}-t_{r}$	$\Delta t_{r}$	0.019 s	0.007	−0.003	−0.018	−0.024
multiple of dipping reflection	$t_{m}^{d}$	1.549 s	1.488	1.437	1.397	1.366
multiple reflection without dip	$t_{m}$	1.510 s	1.498	1.492	1.498	1.510
$t_{m}^{d}-t_{m}.$	$\Delta t_{m}$	−0.039 s	−0.010	0.055	0.101	0.144

Thus the changes in both primaries $\Delta t_{r}$ and multiples $\Delta t_{m}$ are significant. To get the apparent dip of the multiple, we use the data for $x=800\ {\rm {m}}$ in Table 6.10c; the time difference is $(1.549-1.366)=0.183\ {\rm {s}}$. The apparent dip is given by equation (4.2b), assuming ${\bar {V_{2}}}=2950$ so

${\begin{aligned}{\rm {\;sin\;}}\xi _{a}=(2.95/2)(0.183/0.80)=0.337,\xi _{a}=19.7^{\circ }.\end{aligned}}$

The apparent dip moveout for the deep horizontal reflector when the shallow horizon dips $10^{\circ }$ is $1.519-1.471=0.048\ {\rm {s}}$ for $2\Delta x=1600\ {\rm {m}}$; hence it now has the apparent dip given by

${\begin{aligned}{\rm {\;sin\;}}\xi =(2.95/2)(0.048/0.80)=0.089,\ {\rm {or}}\ \xi =5.1^{\circ }.\end{aligned}}$

Figure 6.10c.  Distances for multiples.

6.11 Suppressing multiples by NMO differences

A primary and a multiple each arrive at 0.600 s at $x=0$; their stacking velocities are 1800 and 1500 m/s, respectively. Calculate the residual NMO (after NMO correction for the primary velocity) for offsets of 300 $n$, where $n=1,2,...$. What is the shortest offset that will give good multiple suppression for a wavelet with a 50-ms dominant period?

Solution

The distance to the primary reflector is $(0.600\times 1800)/2=540\ {\rm {m}}$ and to the reflector responsible for the multiple, assuming it is simply a double bounce, is $(0.600\times 1500)/4=225\ {\rm {m}}$. NMO is given by equation (4.1c), $\Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}$. We obtain the following values for the moveouts:

Offset	300 m	600 m	900 m
Primary NMO	0.023 s	0.093 s	0.208 s
Multiple NMO	0.033 s	0.133 s	0.300 s
NMO Difference	0.010 s	0.040 s	0.092 s

Multiple suppression should be maximum when the NMO difference approximates half the wavelet period so that some of the traces are out-of-phase, which is achieved at offset $x$ where

${\begin{aligned}x^{2}/1.200\times 1500^{2}-x^{2}/1.200\times \;1800^{2}\;=0.050,\;x=660\ {\rm {m}}.\end{aligned}}$

6.12 Distinguishing horizontal/vertical discontinuities

Pautsch (1927) showed that a horizontal or vertical interface could give identical first-arrival curves (Figure 6.12a). Add secondary refractions and reflections to show how they can distinguish between the two cases.

Background

At Pautsch’s time only first arrivals were observed and hence interpretation had to be based on them. Today we also observe refraction events (secondary arrivals or secondary refractions) that are not first arrivals (see problem 6.20).

Solution

For the vertical interface in Figure 6.12a(ii) there will be a direct wave plus a reflection directed back toward the source. For the horizontal interface, there is also a reflection and a refraction (head wave), the refraction curve being tangent to the reflection curve at the critical distance. These additional curves distinguish between the two cases. Also, if we move the source, the bend in the curve moves in (ii) but not in (i).

Figure 6.12a.  Models that give identical traveltime curves.

Figure 6.12b.  Models that give identical traveltime curves.

6.13 Identification of events

Figure 6.13a shows events from a high-velocity layer 1.5 wave-lengths thick embedded in lower velocity media; they have been corrected for the normal moveout of the reflection from the top of the layer. Discuss the events and their characteristics.

Background

A wide-angle reflection is one reflected at an angle greater than the critical angle.

Solution

By inspection of Figure 6.13a we note that the embedded wavelet (see Sheriff and Geldart, 1995, p. 284) is approximately symmetrical (zero phase, see Sheriff and Geldart, 1995, p. 553) and apparently has SEG standard polarity (see Sheriff and Geldart, 1995, Figure 6.49) with a central peak for positive reflectivity.

Figure 6.13a.  Source gather (Sheriff and Geldart, 1995, 173) after NMO correction.

The reflection from the top of the layer first decreases in amplitude with offset until the vicinity of $\theta _{c}$ is reached, then it increases in amplitude and becomes a wide-angle reflection and the head wave peels off. The phase of the wide-angle reflection begins to change beyond the critical angle and finally is $180^{\circ }$ out-of-phase with the zero-offset reflection.

The head wave has about the same waveshape as the subcritical reflection and it falls off in amplitude rather rapidly. The reflection from the base of the layer is a negative reflection. It converges on the reflection from the top as the offset increases and its raypath in the high-velocity layer lengthens. Its normal moveout is not hyperbolic. It contributes to the amplitude and phase changes in the reflection from the top of the layer as the two converge.

The converted reflection from the base of the layer involves S-wave travel on either the down-going $({\rm {P}}_{1}{\rm {S}}_{2}{\rm {P}}_{2}{\rm {P}}_{1})$ or up-going $({\rm {P}}_{1}{\rm {P}}_{2}{\rm {S}}_{2}{\rm {P}}_{1})$ legs. They have zero amplitude at zero offset and increase in amplitude with offset; they have the same traveltime and polarity and so reinforce each other.

The converted head wave travels along the interface at the S-wave velocity in the high-velocity layer.

The unidentified event and an associated head wave that project back to zero offset at about 0.71 s may be a reflection from the base of the plate that converts at the top of the plate and travels as an S-wave for both legs in the layer $({\rm {P}}_{1}{\rm {S}}_{2}{\rm {S}}_{2}{\rm {P}}_{1})$ and a head wave that it generates. These would have zero amplitude at zero offset and be weaker than the converted reflection referred to above.

6.14 Traveltime curves for various events

Draw arrival-time curves for the five events in Figure 6.14a.

Figure 6.14a.  Events involving a mesa.

Figure 6.14b.  Arrival times of events. (i) Geometry; (ii) traveltimes.

Solution

We have for the depth to the mesa, 1900 m; height of mesa, 900 m. The traveltime curves were obtained graphically. We let $R$ stand for receiver locations.

For the reflected diffraction from $S_{1}$ (diffracted at A), the virtual source (see problem 4.1) for the event is $I_{1}$ in Figure 6.14b(i) (note that traveltime increases upward), so that

${\begin{aligned}t=(S_{1}A+I_{1}R)/V_{1}=(2.20+I_{1}R)/2.00.\end{aligned}}$

For the reflection from $S_{2}$, we use the virtual source $I_{2}$. We will also have a diffraction from the $S_{2}$ source (paths not shown).

For the reflected refraction from $S_{3}$ (reflected at C), we find two traveltimes and then draw a straight line through them.

${\begin{aligned}{\hbox{At}}\ S_{3},t=2(2.20/2.00+2.20/3.64)=2.80{\hbox{s}}.\\{\hbox{At}}\ S_{4},t=2(2.20/2.00)+1.10/3.64=2.50{\hbox{s}}.\end{aligned}}$

For the diffraction at $C$ from $S_{4}$,

${\begin{aligned}t=(S_{4}C+CR)/2.00=(2.20+CR)/2.00.\end{aligned}}$

For the diffracted reflection from $S_{5}$ (diffracted at C), we use the image point of $S_{5}$ (not shown) so that

${\begin{aligned}t=(I_{5}C+CR)/2.00,\end{aligned}}$

which gives the same curve as for the diffraction from $S_{4}$ except that it is displaced towards increased time by the difference in traveltimes for $S_{4}C$ and $I_{5}C$.

6.15 Reflections/diffractions from refractor terminations

6.15a A horizontal refractor is located under a north-south seismic line at a depth of 1200 m. The overburden velocity is 2500 m/s and the refractor velocity is 4000 m/s. The refractor is terminated by a linear vertical fault (HF in Figure 6.15a) 3500 m from the source point. Determine the traveltime curves when the fault strikes: (i) east-west, (ii) north-south, (iii) N30$^{\circ }$W.

Background

The traveltime curve for a horizontal refractor is given by equation (4.18a). The critical angle $\theta _{c}={\rm {sin}}^{-1}(V_{1}/V_{2})$. In Figure 6.15a the refraction does not exist in the interval SQ; the distance SQ is the critical distance $x^{\prime }$ where

${\begin{aligned}x^{\prime }=2h{\rm {\;tan\;}}\theta _{c}.\end{aligned}}$

(6.15a)

Figure 6.15a.  Events involving refractor termination.

Solution

We require the values of $\theta _{c}$ and $x^{\prime }/2$:

${\begin{aligned}\theta _{c}={\rm {sin}}^{-1}(2.50/4.00)=38.7^{0};x^{\prime }/2=1200{\rm {\;tan\;}}38.7^{0}=0.96\ {\rm {km}}.\end{aligned}}$

The fault is located at $H$ which is more than 960 m from the source in all three cases, so refracted waves are involved.

Case (i). Fault perpendicular to the north-south line.

Events are the following (refer to Figure 6.15a):

1. the direct wave; a straight line through the source with slope (1/2.50) s/km;
2. an inline refraction, a straight line beginning at $Q$ with slope (1/4.00) s/km; it is tangent at $Q$ to the reflection hyperbola (curve 5);
3. a reflected head wave between $S$ and $G$ from reflection point $H$ with paths such as $STHMP$, a straight line extending from $S$ to $G$ with slope opposite to (2) and a larger intercept time;
4. a diffraction generated at $H$, a curve through $F$ symmetrical about the vertical and tangent to the head-wave curve (2) at $U$ (if prolonged beyond $F$) and to curve (3) at $G$;
5. a reflection, a hyperbola symmetrical about the vertical through $S$ and tangent to curve (2) at $Q$;
6. a reflected refraction such as $STJKL$ (if the impedance contrast at the fault is large enough to produce a recognizable reflection), a straight line extension of curve (3) to the right of $G$.

Case (ii). Fault parallel to the seismic line.

A plan view of the fault is shown to the left of the “line” in Figure 6.15b. The observed events are

1. the direct wave; a straight line passing through $S$ with slope (1/2.50) s/km;
2. an inline refraction;
3. a reflected refraction along paths such as STFMP shown in plan view in Figure 6.15b where the energy goes down from $S$ at the critical angle until the refractor is reached at $T$, then along $TF$ to the fault at $F$ where reflection occurs, after which the energy travels along $FM$ until a ray peels off at $M$ and travels up to the recorder at $P$, its traveltime curve is shown in Figure 6.15c; the curve is given by equation (4.18a) where we replace $x$ with the distance

   ${\begin{aligned}SF+FP=2SF=2[y^{2}+(x/2)^{2}]^{1/2}=2[3.50^{2}+x^{2}/4]^{1/2},\end{aligned}}$

   $x$ here being the distance $SP$; the curve is a hyperbola (see Figure 6.15c);

4. no diffractions will be observed because there is no point source;
5. inline reflections; the reflection curves are the usual hyperbolas;
6. there is no reflected refraction such as STJKL in Figure 6.15a in case (i) because it could only exist within a distance of 0.96 km from the fault.

Figure 6.15b.  Travelpaths for faults parallel and at angle to seismic line.

Figure 6.15c.  Arrival time curve for the reflected refractions.

Case (iii). Fault at an angle to the seismic line.

This case is similar to case (ii). The fault in the plan view shown to the right of the “line” in Figure 6.15b strikes at the angle N30$^{\circ }$W. The observed events are

1. the direct wave;
2. an inline refraction;
3. a reflected refraction, a typical path being ${ST}^{'}{F}^{'}{M}^{'}{P}$,$T^{'}$ and $M^{'}$ being equivalents of $T$ and $M$ in Figure 6.15b; its traveltime curve is shown in Figure 6.15c; to derive the traveltime curve for this event, we use image point $I$ for reflection in the fault, so $IP$ replaces $x$ in equation (4.18a). We get

   ${\begin{aligned}t=IP/V_{2}+(2h{\rm {\;cos\;}}\theta _{c})/V_{1}\\=x^{2}+SI^{2}-2xSI{\rm {\;cos\;}}60^{\circ })^{1/2}/4.00+2\times 1.20{\rm {\;cos\;}}38.7^{\circ }/2.50\\=(1/4.00)(x^{2}-7.00x+49.00)^{1/2}+0.749.\end{aligned}}$

   Thus, $(4-2.96)^{2}=(x^{2}-7.00x+49.00)$. The curve is a hyperbola (see Figure 6.15c);

4. no diffraction event;
5. a normal reflection;
6. same as in case (ii).

All traveltime curves are normal except (3).

6.15b Repeat for the east-west fault for a refractor that dips $10^{\circ }$ to the north with the source to the south.

Solution

In Figure 6.15d, $\theta _{c}=38.7^{\circ }$, $SB=1.20$ km, $BH=3.50$ km. We make frequent use of the law of sines to calculate distances:

${\begin{aligned}BT=(1.20/{\rm {\;sin\;}}51.3^{\circ }){\rm {\;sin\;}}38.7^{\circ }=0.74\ {\rm {km}},\\ST=1.2{\rm {\;sin\;}}100^{\circ }/{\rm {\;sin\;}}51.3^{\circ }=1.51\ {\rm {km}},\\TH=QC=3.50-BT=2.76\ {\rm {km}},\\SQ=ST{\rm {\;sin\;}}77.4^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=2.23\ {\rm {km}},\\TQ=ST{\rm {\;sin\;}}61.3^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=2.01\ {\rm {km}},\\QU=QC{\rm {\;sin\;}}128.7^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=3.26\ {\rm {km}},\\SU=SQ+QU=5.49\ {\rm {km}},\\CU=QC{\rm {\;sin\;}}10^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=0.73\ {\rm {km}},\\HU=HC+CU=TQ+CU=2.74\ {\rm {km}},\\HG=HU{\rm {\;sin\;}}41.3^{\circ }/{\rm {\;sin\;}}61.3^{\circ }=2.06\ {\rm {km}},\\HD=1.20+3.50{\rm {\;sin\;}}10^{\circ }=1.81\ {\rm {km}},\\SG=SU-GU=SU-(HU{\rm {\;sin\;}}77.4^{\circ }/{\rm {\;sin\;}}61.3^{\circ })\\=5.49-3.05=2.44\ {\rm {km}},\\SD=3.50{\rm {\;cos\;}}10^{\circ }=3.45\ {\rm {km}}.\end{aligned}}$

Figure 6.15d.  Events involving termination of a dipping refractor ($\theta =$ critical angle).

For the refraction,

${\begin{aligned}t_{Q}=(ST-TQ)/2.50=1.41\ {\rm {s}},\\t_{U}=(ST+HU)/2.50+TH/4.00=2.39\ {\rm {s}}.\end{aligned}}$

For the reflected refraction,

${\begin{aligned}t_{G}=(ST+HG)/2.50+TH/4.00=2.12\ {\rm {s}},\\t_{S}=2(ST/2.50+TH/4.00)=2.59\ {\rm {s}}.\end{aligned}}$

The diffraction from $H$ has its minimum traveltime at $D$:

${\begin{aligned}t_{D}=(ST+HD)/2.50+TH/4.00=2.02\ {\rm {s}}.\end{aligned}}$

For the reflection,

${\begin{aligned}t_{0}=2(1.20{\rm {\;cos\;}}10^{\circ })/2.50=0.95\ {\rm {s}},\\t_{Q}={\rm {same\ as}}\ t_{Q}\ {\rm {for\ the\ refraction}}\ =1.41\ {\rm {s}}.\end{aligned}}$

6.15c What effect will the manner of terminating the refractor have, that is, how will the amplitude of the reflected refraction depend on the dip of the terminating fault?

Solution

Provided the impedance contrast across the fault is large enough, any abrupt termination of the refractor will generate a reflected refraction. The attitude of the terminating fault will have a relatively small effect on the amplitude provided that the dip of the fault is such that the angle of incidence is close to $90^{\circ }$.

6.15d Most commonly a faulted refractor terminates against rock of lower acoustic impedance, but the opposite situation can also happen. What differences will this make?

Solution

The nature of refractors is that they have high velocity, hence usually terminate against rocks of lower acoustic impedance and a reflected refraction has opposite phase to the incident refraction. However, if a refractor terminates against a higher impedance, they will have the same phase.

6.15e Extend the profile for part (a), case (i), an appreciable distance beyond the fault so as to plot the diffraction from the refractor termination. Assume uniform 2.50-km/s material beyond the refractor termination.

Solution

The extensions of curves (1), (2), (4), and (5) are shown in Figure 6.15a. Events (3) and (6) do not exist to the right of the fault.

Figure 6.16a.  Refraction multiple.

6.16 Refractions and refraction multiples

6.16a Determine the traveltime curve for the refraction ${\mathit {SMNPQR}}$ and the refraction multiple ${\mathit {SMNTUWPQR}}$ in Figure 6.16a.

Solution

We assume that the velocities are known to three significant figures. Then, using equation (3.1a),

${\begin{aligned}\theta _{2}={\rm {sin}}^{-1}(2.00/4.20)=28.4^{0};\theta _{1}={\rm {sin}}^{-1}(2.80/4.20)=41.8^{\circ }\end{aligned}}$

The traveltimes can be obtained either graphically or by calculation. Calculating, we get for the refraction traveltimes $t_{R}$

${\begin{aligned}t_{R}=2SM/2.80+2MN/2.00+NP/4.20\\=2\times 0.75/2.80{\rm {\;cos\;}}41.8^{\circ }+2\times 3.25/(2.00{\rm {\;cos\;}}28.4^{\circ })\\+(x-2\times 0.75{\rm {\;tan\;}}41.8^{\circ }-2\times 3.25{\rm {\;tan\;}}28.3^{\circ })/4.20\\=x/4.20+2\times 0.75{\rm {\;cos\;}}41.8^{\circ }/2.80+2\times 3.25{\rm {\;cos\;}}28.4^{\circ }/2.00\\=x/4.20+3.26.\end{aligned}}$

The critical distance (see equation (6.15a) is

${\begin{aligned}x^{'}=2(0.75{\rm {\;tan\;}}41.8^{\circ }+3.25{\rm {\;tan\;}}28.4^{\circ })=4.86\ {\rm {km}}.\end{aligned}}$

The traveltime curve for SMNTUWPQR is parallel to that for SMNPQR and displaced toward longer time by the amount $\Delta t$ where

${\begin{aligned}\Delta t=2TU/2.00-TW/4.20\\=2\times 3.25/(2.00{\rm {\;cos\;}}28.4^{\circ })-2\times 3.25({\rm {\;tan\;}}28.4^{\circ })/4.20=2.86\ {\rm {s}}.\end{aligned}}$

The critical distance for SMNTUWPQR is increased to

${\begin{aligned}x^{'}=2\times 0.75{\rm {\;tan\;}}41.8^{\circ }+4\times 3.25{\rm {\;tan\;}}28.4^{\circ }=8.37\ {\rm {km}}.\end{aligned}}$

The traveltime curves are plotted as curves (a) in Figure 6.16b.

Figure 6.16b.  Traveltime curves. Letters denote curves for respective parts (a), (b), (c).

6.16b Determine the traveltime curves when both refractor and reflector dip $8^{\circ }$ down to the left, the depths shown in Figure 6.16a now being the slant distances from $S$ to the interfaces.

Solution

A combined graphical and calculated solution probably provides the easiest solution although Adachi’s method (see problem 11.5) could be used to give greater precision if the data accuracy warranted. A large-scale graph was used to achieve better accuracy; Figure 6.16c is a reduced-scale replica. The traveltime curves are shown in Figure 6.16b labeled (b).

The critical distance for the refraction is $SR_{1}=4.38$ km, and

${\begin{aligned}t_{R_{1}}=(1.00+0.18)/2.80+2\times 3.71/2.00=4.13\ {\rm {s}}.\end{aligned}}$

The $V_{2}$-layer outcrops at $R_{2}=5.39$ km,

${\begin{aligned}t_{R_{2}}=1.00/2.80+2\times 3.71/2.00+1.12/4.20=4.33\ {\rm {s}}.\end{aligned}}$

Figure 6.16c.  Geometry and raypaths for dip $8^{\circ }$ to the left.

The headwave has a different slope to the right of $R_{2}$. To plot the curve in this zone, we use point $R_{4}$ at the offset $x=8.79\ {\rm {km}}$. Then,

${\begin{aligned}t_{R_{4}}=1.00/2.80+(3.71+3.21)/2.00+4.71/4.20=4.94\ {\rm {s}},\end{aligned}}$

and the headwave curve is a straight line joining the traveltimes at the points $R_{2}$ and $R_{4}$.

We have two types of reflected refractions: a typical path for the first type is $SMNP_{1}U_{2}WP_{4}R_{4}$, the reflection occurring at the shallow dipping interface, The second type, $SMNP_{5}R_{6}P_{7}R_{7}$, involves reflection at the surface. The first type exists between $R_{3}$ and $R_{4}$, and the curve is parallel to the head-wave curve to the right of $R_{3}$. The second type exists to the right of $R_{4}$ and the curve is parallel to the other reflected refraction. To plot the reflected-refraction curves, we need one point on each curve and then use the refraction-curve slope to the right of $R_{3}$. For the first type, we find the coordinates of $R_{5}$:

${\begin{aligned}x=6.80\ {\hbox{km}},\\t_{R_{5}}=1.00/2.80+(2\times 3.71+3.32+3.12)/2.00+1.12/4.20=7.55\ {\rm {s}}.\end{aligned}}$

For the second type we find coordinates of $R_{7}$:

${\begin{aligned}x=9.15\ {\rm {km}},\\t_{\rm {R_{7}}}=1.00/2.80+(3.71+3.30+2.97+2.78)/2.00+3.85/4.20=7.65\ {\rm {s}}.\end{aligned}}$

6.16c What happens when the reflector dips $3^{\circ }$ to the left and the refractor $5^{\circ }$to the left?

Figure 6.16d.  Geometry and raypaths for $3^{\circ }$ and $5^{\circ }$ dips to the left.

Solution

A summary of the detailed graphical solution is as follows. The traveltime curves are shown in Figure 6.16b labeled (c). The various angles in Figure 6.16d are

${\begin{aligned}\theta _{c}=28.4^{\circ }\;,\;\theta _{2}=(28.4^{\circ }+5^{\circ }-3^{\circ })\;=30.4^{\circ },\\\theta _{1}={\rm {sin}}^{-1}[(2.80/2.00){\rm {\;sin\;}}30.4^{\circ }]=45.1^{\circ },\\\alpha _{1}={\rm {angle\ of\ approach}}\ =45.1^{\circ }+3^{\circ }=48.1^{\circ },\\\theta _{2}^{'}=(28.4^{\circ }-5^{\circ }+3^{\circ })\;=26.4^{\circ },\\\theta _{1}^{'}={\rm {sin}}^{-1}[(2.80/2.00){\rm {\;sin\;}}38.5^{\circ }]=38.5^{\circ },\\\alpha _{1^{'}}=38.5^{\circ }-3^{\circ }\;=35.5^{\circ }\end{aligned}}$

The refraction curve is a straight line though $R_{5}$ and $R_{7}$:

${\begin{aligned}R_{5}:x^{'}=4.64\ {\rm {km}}={\rm {critical\ distance}},\\t_{R_{5}}=(1.09+0.44)/2.80+(3.65+3.50)/2.00=4.20\ {\rm {km}};\\R_{7}:x=8.15\ {\rm {km}},\\t_{R_{7}}=(1.09+0.44)/2.80+(3.64+3.35)/2.00+3.80/4.20\\=4.95\ {\rm {s}}.\end{aligned}}$

The incident angle at $W$ is $\theta _{3}=24.4^{\circ }$, which is less than the critical angle, so that no refraction will be generated there, only a reflection. However, the refraction that starts at $N$ will give rise to upgoing rays which will be reflected, giving a reflected refraction, such as the ray that ends at $R_{8}$. The traveltime curve is parallel to the refraction curve and exists beyond $R_{6}$ whose coordinates are

${\begin{aligned}x=7.79\ {\rm {km}},\\t_{R_{6}}=(1.09+0.48)/2.80+(3.64+3.50+3.42+3.36)/2.00=7.52\ {\rm {s}}.\end{aligned}}$

Figure 6.17a.  Reflections where the second and third reflectors converge; zero-phase wavelet.

6.17 Destructive and constructive interference for a wedge

Figures 6.17a show three reflections, where the second and third reflections are from the top and bottom of wedges that converge to the right. Explain why waves in Figure 6.17a(i) interfere destructively and in Figure 6.17a(ii) constructively when the wedge thickness is ${\frac {1}{4}}\lambda$.

Solution

In Figure 6.17a(i) both reflections from the wedge have the same polarity. As the reflectors converge, at a thickness of ${\frac {1}{4}}\lambda$ (2-way distance ${\frac {1}{4}}\lambda$) one half-cycle of the wavelet reflected from the base interferes destructively with the next half-cycle from the top. In Figure 6.17a(ii), where a phase reversal occurs on reflection at one surface but not at the other surface, the reflections from the top and base of the wedge interfere constructively at ${\frac {1}{4}}\lambda$ thickness (before they undergo destructive interference as they converge further). Note that timing the peaks or troughs does not give the correct traveltimes to the respective interfaces where the thickness is $<{\frac {1}{4}}\lambda$ (about 8 ms in Figure 6.17a).

6.18 Dependence of resolvable limit on frequency

6.18a A wavelet has a flat frequency spectrum from 0 to $f_{u}$ above which no frequencies are present. Show that the Rayleigh criterion gives a resolvable limit $t_{r}$, where $t_{r}=0.715/f_{u}$.

Figure 6.18a.  Illustrating resolution.

Background

A reflecting layer is said to be resolvable if we can distinguish between the reflections from the top and bottom of the layer, usually on the basis of a phase break in the superimposed reflections (see Figure 6.18a). The Rayleigh criterion for vertical resolution states that at least a small depression must appear between successive events in order to recognize that more than one event is present. For this to occur the two reflections must be separated by at least a half-cycle; this corresponds to a minimum thickness of $\lambda /4$, the two-way thickness then being $\lambda /2$. This thickness of ${\frac {1}{4}}\lambda$ is called the tuning thickness or resolvable limit.

Figure 6.18b.  Boxcar and its transform.

A boxcar (see Figure 6.18b) is a function whose value is unity within a certain range and zero outside this range (see problem 9.3).

Solution

The frequency spectrum is a boxcar extending from $-f_{u}\ {\rm {to}}+f_{u}$, shown in Figure 6.18b, which we write as ${\rm {box}}_{2f_{u}}(f)$. The inverse transform of the spectrum is given by equation (9.3d), namely

${\begin{aligned}g(t)=(1/2\pi )\mathop {\int } \nolimits _{-\infty }^{+\infty }\ {\rm {box}}_{2f_{u}}(f)e^{j\omega t}\ {\rm {d}}\omega =\mathop {\int } \nolimits _{-f_{u}}^{+f_{u}}e^{j2\pi ft}\ {\rm {d}}f\\=(1/2\pi jt)(e^{j2\pi f_{u}{f}}-e^{-j2\pi f_{u}{f}})=(1/\pi t){\rm {\;sin\;}}(2\pi f_{u}t)\\=2f_{u}\ {\rm {sinc}}\ (2\pi f_{u}t),\end{aligned}}$

(6.18a)

where sinc $x=({\rm {\;sin\;}}x)/x.$

The Rayleigh criterion gives a resolvable limit $t_{r}$ corresponding to the first trough (minimum) of the time-domain representation of the boxcar. Hence, we equate the derivative of the sinc function to zero, obtaining

${\begin{aligned}{\frac {\rm {d}}{{\rm {d}}t}}[{\rm {sinc}}(2\pi f_{u}t)]={\frac {\rm {d}}{{\rm {d}}t}}\left[{\frac {{\rm {\;sin\;}}(2\pi f_{u}t)}{2\pi f_{u}t}}\right]=0={\frac {\rm {d}}{{\rm {d}}t}}\left[{\frac {{\rm {\;sin\;}}(2\pi f_{u}t)}{t}}\right],\end{aligned}}$

${\begin{aligned}{\mbox{so}}\qquad \qquad \ (2\pi f_{u}/t){\rm {\;cos\;}}(2\pi f_{u}t)-(1/t^{2}){\rm {\;sin\;}}(2\pi f_{u}t)=0;\end{aligned}}$

This gives ${\rm {\;tan\;}}(2\pi f_{u}t)=2\pi f_{u}t$, that is, we must solve the equation ${\rm {\;tan\;}}x=x$ where $x=2\pi f_{u}t$. A graphical solution gives $x=4.49$, hence $t_{r}=0.715/f_{u}$.

6.18b Show that the value of $t_{r}$ for a wavelet with a flat spectrum extending from $f_{L}$ to $nf_{L}$ (that is, $m$ octaves where $n=2^{m}$) is given by the solution of the equation,

${\begin{aligned}nx\ {\rm {cos}}\ nx-{\rm {\;sin\;}}nx\\-x{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0,\end{aligned}}$

where $x=2\pi f_{L}t_{r}$.

Figure 6.18c.  Boxcar spectrum.

Solution

The time-domain function corresponding to the spectrum in Figure 6.18c is

${\begin{aligned}g(t)=(1/2\pi )\left(\mathop {\int } \nolimits _{-nf_{L}}^{-f_{L}}e^{1^{\omega t}}\ {\rm {d}}\omega +\mathop {\int } \limits _{f^{L}}^{nf_{L}}e^{j\omega t}\ {\rm {d}}\omega \right)\\=(1/2\pi jt)[(e^{-j2\pi f_{L}{t}})-(e^{-j2\pi nf_{L}{t}})+(e^{j2\pi f_{L}{t}}-e^{j2\pi nf_{L}{f}})]\\=(1/\pi t)[{\rm {\;sin\;}}(2\pi nf_{L}t)-{\rm {\;sin\;}}(2\pi f_{L}t)].\end{aligned}}$

To get the first trough, we write $x=2\pi f_{L}t$, then equate to zero the derivative of $g(x)$ with respect to $x$. This gives

${\begin{aligned}g(x)=(2f_{L}/x)({\rm {\;sin\;}}nx-{\rm {\;sin\;}}x),\\dg(x)/dx=0=2f_{L}[(n{\rm {\;cos\;}}nx-{\rm {\;cos\;}}x)/x-({\rm {\;sin\;}}nx-{\rm {\;sin\;}}x)/x^{2}],\\{\mbox{so}}\qquad \qquad \ nx{\rm {\;cos\;}}nx-{\rm {\;sin\;}}nx-x{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0.\end{aligned}}$

(6.18c)

6.18c Solve the equation in part (b) for $m=3,2,1.5,$ and 1, that is, for bandwidths of 3, 2, 1.5, and 1 octaves, and compare the relation between $t_{r}$ and $m$.

Solution

For $m=3$, $n=2^{3}=8$ and we have from equation (6.18c)

${\begin{aligned}8x{\rm {\;cos\;}}8x-{\rm {\;sin\;}}8x-{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0.\end{aligned}}$

For $x=0$, we get $0=0$, but for $x$ slightly greater than zero, ${\rm {d}}g(x)/{\rm {d}}x$ is negative and continues to be negative until it changes sign for $x$ between 0.5 and 0.6; The corresponding root is $x=0.560$, giving $t_{r}=0.089/f_{L}.$.

When $m=2$, $n=4$, the root of equation (6.18c) is $x=1.10$, giving $t_{r}=0.175/f_{L}$. For $m=1.5$, $n=2{\sqrt {2}}=2.83$and the root is $x=1.51$, so $t_{r}=0.240/f_{L}$. For $m=1.0$, $n=2$ and the root is $x=2.015$, so $t_{r}=0.321/f_{L}$.

6.18d Noting that part (a) involves an infinite number of octaves, what bandwidth is required to give nearly the same result?

Solution

The resolution in part (a) is expressed in terms of $f_{u}$ while those in (c) are in terms of $f_{L}$. To compare the results we equate $nf_{L}$ to $f_{u}$, so that we have four frequency bands, each with top frequency $f_{u}$ and extending downward 1, 1.5, 2, 3, and an infinite number of octaves. This means that the values of $t_{r}$ in (c) must be adjusted to get $nf_{L}$ in the denominator, e.g.,for $n=4$, $t_{r}=4\times 0.175/(4f_{L})$. The results are shown in Table 6.18a.

Table 6.18a. Resolution versus number of octaves.

$m$	$n$	$t_{r}$/($nf_{L}$)
$\infty$	$\infty$	0.715
3	8	0.712
2	4	0.700
1.5	2.83	0.679
1	2	0.321

Thus, three octaves bandwidth ($m=3$) gives almost as good resolution as an infinite number of octaves but the resolution deteriorates for band-widths $<1.5$ octaves.

6.19 Vertical resolution

6.19a Approximately, what are the dominant frequencies for reflections in Figure 6.19a that arrive at the right side of the section at about 0.6, 1.2, and 1.8 s?

Background

Resolution is discussed in problem 6.18 where it is shown that the resolvable thickness is $\lambda /4$.

Solution

Measuring peak-to-peak time intervals for several cycles on an enlarged figure, we get periods of about 20, 30, and 40 ms, or frequencies of 50, 33, and 25 Hz, for the three reflections.

Figure 6.19a.  Section in Ardmore Basin, Oklahoma.

6.19b If the velocities at these reflectors are 2.0, 3.0, and 5.0 km/s, respectively, what are the resolvable thicknesses?

Solution

Resolvable thicknesses are about $\lambda /4=V/4f$. This yields thicknesses of 10, 23, and 50 m, respectively.

6.20 Causes of high-frequency losses

Denham’s (1980) empirical high-frequency limit is related both to the loss of high frequencies and to the dynamic range of the recording system. Reconcile this limit with losses by absorption of $0.15\ {\rm {dB}}/\lambda$, spreading, and high-frequency loss because of peg-leg multiples (as illustrated in Figures 6.20a,b). Take 84 dB as the dynamic range of the recording-system.

Figure 6.20a.  Peg-leg mechanism.

Background

Denham’s high-frequency limit is discussed in problem 7.4 and absorption in problem 2.18. Absorption versus spreading (divergence) is the subject of problem 3.8. Spreading causes the amplitude to fall off inversely with distance. Peg-leg multiples (illustrated in Figure 6.20a,b) are described in problem 3.8.

The dynamic range of a recording system is the ratio (usually expressed in decibels) of the maximum signal that can be amplified without distortion to the background noise of the system.

Solution

Denham’s relation is $f_{\rm {\;max\;}}=150/t$, where $f_{\rm {\;max\;}}$is the maximum usable frequency and $t$ is the traveltime. We assume two depths 100 and 4100 m and find the average velocity ${\bar {V}}$ for this depth interval using the South Louisiana curve in Figure 5.5b; the result is ${\bar {V}}=2600\ {\rm {m/s}}$. Thus, $t\approx 3.1\ {\rm {s}}$ and Denham’s relation gives $f_{\rm {\;max\;}}=48$ Hz.

Figure 6.20b.  Attenuation because of peg-legs (from Schoenberger and Levin, 1978).

For a frequency of 48 Hz,$\lambda =2600/48=54\ {\rm {m}}$. The total distance traveled by the wave $=8000\ {\rm {m}}=148\lambda$. From Figure 6.20b, we take $0.08\ {\rm {dB}}/\lambda$ as a median value of $\eta \lambda$ for the loss due to peg-leg multiples, the total loss being $0.08\times 148=12\ {\rm {dB}}$. Taking absorption losses as $0.15\ {\rm {dB}}/\lambda$ gives 22 dB for absorption. Spreading causes the amplitude to decrease inversely as the distance and the distance ratio is 4100/100 = 41, giving a loss of $20{\rm {\;log\;}}41=32\ {\rm {dB}}$. Thus, of the total losses of 66 dB, about 20% is due to peg-leg multiples, 30% to absorption, and 50% to spreading. Before generalizing these conclusions, we must realize that spreading losses decrease rapidly with distance compared to peg-leg multiples and absorption losses.

Figure 6.21a.  Ricker wavelet (i) in time domain and (ii) in frequency domain.

The total loss of 66 dB over the 4000 m interval is well within the 84 dB dynamic range of the system.

6.21 Ricker wavelet relations

6.21a Verify that the Ricker wavelet in Figure 6.21a(i),

${\begin{aligned}g(t)=(1-2\pi ^{2}f_{m}^{2}t^{2})e^{-(\pi f_{m}t)^{2}},\end{aligned}}$

(6.21a)

$f_{m}$, being the peak frequency, has the Fourier transform [Figure 6.21a(ii)]

${\begin{aligned}G(f)=(2/{\sqrt {\pi }})(f/f_{m})^{2}e^{-(f/f_{m})^{2}},\;\gamma (f)=0,\end{aligned}}$

(6.21b)

where $\gamma (f)$ is the phase.

Background

Fourier transforms are discussed in problem 9.3 and theorems on Fourier transforms in Sheriff and Geldart, 1995, section 15.2.6.

The transform of $e^{-at^{2}}$ is

${\begin{aligned}e^{-at^{2}}\leftrightarrow (\pi /a)^{1/2}e^{-\omega ^{2}/4a}\end{aligned}}$

(6.21c)

[Papoulis, 1962: p. 25, equation (2-68)].

Solution

The time-domain expression for the Ricker wavelet can be written in the form

${\begin{aligned}g(t)=(1-2at^{2})e^{-at^{2}}=e^{-at^{2}}-2at^{2}e^{-at^{2}},\end{aligned}}$

(6.21d)

where $a=(\pi f_{M})^{2}$. The transform of the first term is $(\pi /a)^{1/2}e^{-\omega ^{2}}/4a$. To get the transform of the second term, we use Sheriff and Geldart, 1995, equation (15.142) which states that when $g(t)\leftrightarrow G(\omega )$, then,

${\begin{aligned}(-jt)^{n}g(t)\leftrightarrow {\frac {{\rm {d}}^{n}G(\omega )}{{\rm {d}}\omega ^{n}}},\end{aligned}}$

that is, for $n=2$,

${\begin{aligned}-t^{2}g(t)\leftrightarrow {\frac {{\rm {d}}^{2}G(\omega )}{{\rm {d}}\omega ^{2}}}.\end{aligned}}$

The transform of the second term now becomes

${\begin{aligned}2a{\frac {{\rm {d}}^{2}}{{\rm {d}}\omega ^{2}}}\left[\left({\frac {\pi }{a}}\right)^{1/2}e^{-(\omega ^{2}/4a)}\right]=2(\pi a)^{1/2}{\frac {\rm {d}}{{\rm {d}}\omega }}\left[-\left({\frac {\omega }{2a}}\right)e^{-(\omega ^{2}/4a)}\right]\\=-(\pi /a)^{1/2}{\frac {\rm {d}}{{\rm {d}}\omega }}\left[\omega e^{-(\omega ^{2}/4a)}\right]=-(\pi /a)^{1/2}\left[1-\left({\frac {\omega ^{2}}{2a}}\right)\right]e^{-(\omega ^{2}/4a)}.\end{aligned}}$

Adding the two transforms, we have

${\begin{aligned}&g(t)\leftrightarrow (2/{\sqrt {\pi }})(\omega /\omega _{M})^{2}e^{-(\omega /\omega _{M})^{2}}=(\pi /a^{3})^{1/2}(\omega ^{2}/2)e^{-(\omega /\omega _{M})^{2}}\\&\leftrightarrow (2/{\sqrt {\pi }})(f^{2}/f_{M}^{3})e^{-(f/f_{M})^{2}}.\end{aligned}}$

6.21b Show that $f_{M}$ is the peak of the frequency spectrum.

Solution

To find the peak frequency, we set the derivative ${\rm {d}}G(f)/{\rm {d}}f$ equal to zero. Thus, omitting the constant factor,

${\begin{aligned}\left({\frac {\rm {d}}{{\rm {d}}f}}\right)\left[f^{2}e^{-(f/f_{M})^{2}}\right]=e^{-(f/f_{M})^{2}}\left[2f-f^{2}(2f/f_{M}^{2})\right]=0,\end{aligned}}$

so $f=f_{M}$ for a maximum.

6.21c Show that $T_{D}/T_{R}={\sqrt {3}}$ (see Figure 6.21a) and that $T_{D}f_{M}={\sqrt {6}}/\pi .$

Solution

Since $g(t)=0$ for $t=+T_{R}/2$, we have

${\begin{aligned}[1-2(\pi f_{M}T_{R}/2)^{2}]e^{-(\pi f_{M}T_{R}/2)^{2}}=0,\end{aligned}}$

(6.21e)

hence $T_{R}=\pm {\sqrt {2}}/(\pi f_{M})$.

Moreover, $g(t)$ is a minimum for $t=\pm T_{D}/2$, so $T_{D}/2$ is a root of

${\begin{aligned}({\rm {d}}/{\rm {d}}t)[(1-2at^{2})e^{-at^{2}}]=0,\end{aligned}}$

that is, of the equation

${\begin{aligned}e^{-at^{2}}[-4at+(1-2at^{2})(-2at)]=0,\\{\mbox{so}}\qquad \qquad \ t=T_{D}/2=[{\sqrt {(3/2)}}]/\pi f_{M}.\end{aligned}}$

(6.21f)

Hence, $T_{D}/T_{R}={\sqrt {3}}$ and $T_{D}f_{M}={\sqrt {6/\pi }}$.

6.22 Improvement of signalnoise ratio by stacking

Select random numbers between $\pm 9$ to represent noise $N_{j}$ and add to each a signal $S=2$. Sum 4 values of $S+N_{j}$ and determine the mean, standard deviation $\sigma$ of $(S+N_{j})$, and values of the ratio $S/(S+N)$. Repeat for 8, 16, and 32 values. Note how the mean converges toward $S$ as the number of values increases, how $\sigma$ approaches a limiting value (which depends on the statistical properties of the noise), and how the ratio $S/(S+N)$ converges toward $+1$.

Background

The standard deviation $\sigma$ is a measure of the scatter of measured values of a quantity, large values corresponding to large variations. It is given by the equation

${\begin{aligned}\sigma =[(1/n)\Sigma (x-{\bar {x}})^{2}]^{1/2},\end{aligned}}$

(6.22a)

where $n$ is the number of values and ${\bar {x}}$ is the mean value.

Table 6.22a. Random numbers.

20897	13007	95217	19221	15433	94882	23741	86571	20504	22169
20737	19305	71148	04035	03180	79506	12771	34806	37279	62739
31552	59282	16856	38655	31802	84283	08694	06945	19286	16924
60605	97685	26147	51379	39553	04893	25469	96469	57436	97888
42094	17446	27775	99466	63704	60957	55029	02764	91845	76174
54774	15832	04324	73597	42328	74303	58231	85798	16725	27836
89730	31886	34683	07814	57000	63721	43798	12003	04676	08367
12049	18538	96266	62439	81839	13093	22659	75018	31494	89519
22364	15913	51674	94189	10336	97801	21025	58966	40663	26197
80102	39977	78674	29634	38652	85289	47962	16594	50834	93484

To obtain a sequence of random numbers within a given range, adopt a rule for selecting digits from Table 6.22a; use this rule to get the units digit, then get the tens digit, etc. To determine the algebraic sign of each number, adopt another rule for fixing the sign, e.g., letting even/odd digits signify plus/minus, respectively.

Solution

Using Table 6.22a, we get 32 values of $N_{i}$ and sums $S+N$ in Table 6.22b.

Writing $({\overline {S+N}})$ for the mean of ($S+N_{i}$), we get the results in Table 6.22c.

Table 6.22b. Values of $N_{i}$ and $S+N_{i}$.

$N_{i}$	$S+N_{i}$	$N_{i}$	$S+N_{i}$	$N_{i}$	$S+N_{i}$	$N_{i}$	$S+N_{i}$
$-2$	0	$-2$	0	$-3$	$-1$	$+5$	$+7$
$-1$	$+1$	$+1$	$+3$	$-3$	$-1$	$+8$	$+10$
$+9$	$+11$	$+7$	$+9$	$-8$	$-6$	$-8$	$-6$
0	$+2$	$+2$	$+4$	0	$+2$	$-3$	$-1$
$-1$	$+1$	$-6$	$-4$	0	$+2$	$+5$	$+7$
$+9$	$+11$	$+3$	$+5$	$-4$	$-2$	$-4$	$-2$
$+2$	$+4$	$-5$	$-3$	$+4$	$+6$	$-6$	$-4$
$-8$	$-6$	$-1$	$+1$	$+7$	$+9$	$+1$	$+3$

Table 6.22c. Calculated values of $({\overline {S+N}})$, $\sigma$, and $S({\overline {S+N}})$ for different numbers of samples.

	$\Sigma (S+N)$	$({\overline {S+N}})$	$\sigma$	$S/({\overline {S+N}})$
1st 4	14	3.5	4.4	0.57
1st 8	24	3.0	5.3	0.67
2nd 8	15	1.9	4.0	1.05
3rd 8	9	1.1	4.4	1.82
4th 8	14	1.8	5.5	1.11
1st 16	39	2.4	4.8	0.83
2nd 16	23	1.4	5.0	1.43
32	62	1.9	4.9	1.05

We see that as the number of samples increases, $\sigma$ approaches 4.9, $({\overline {S+N}})$ approaches 2, and $S/({\overline {S+N}})$ approaches $+1$.